Let g(x) = ax + b

f(x) is continuous at x = 0

So ${\mathrm{lt}}_{\mathrm{x}\to 0^{+}}{\left(\frac{1+\mathrm{x}}{2+\mathrm{x}}\right)}^{\frac{1}{\mathrm{x}}}={\left(\frac{1}{2}\right)}^{\mathrm{\infty }}=0$

${\mathrm{\ell }}_{\mathrm{x}\to 0^{-}}^{\mathrm{\ell }\left(\mathrm{x}\right)}\mathrm{g}\underset{\mathrm{x}\to 0^{-}}{\mathrm{\ell t}}\mathrm{ax}+\mathrm{b}=\mathrm{b}$

So b = 0

$\mathrm{Now} \mathrm{f}\left(\mathrm{x}\right)={\left(\frac{1+\mathrm{x}}{2+\mathrm{x}}\right)}^{\frac{1}{\mathrm{x}}}$

ℓnf (x) = $\frac{1}{\mathrm{x}}$ [ℓn( 1 + x) – In(2 + x)] 

$\frac{1}{\mathrm{f}\left(\mathrm{x}\right)}\mathrm{f}\left(\mathrm{x}\right)=\frac{\mathrm{x}\left[\frac{1}{1+\mathrm{x}}-\frac{1}{2+\mathrm{x}}\right]-\ln \left(\frac{1+\mathrm{x}}{2+\mathrm{x}}\right)}{{\mathrm{x}}^2}$

$\mathrm{f}\left(1\right)=\mathrm{f}\left(1\right)\times \left[\frac{\left[\frac{1}{2}-\frac{1}{3}\right]-\ln \frac{2}{3}}{1}\right]$

$\mathrm{f}\left(1\right)=\frac{2}{3}\left[\frac{1}{6}-\ln \frac{2}{3}\right]$

Also f(–1) = (ax + b) at x = –1 = – a

$\text{ So }-\mathrm{a}=\frac{1}{9}-\frac{2}{3}\ln \frac{2}{3}$

$\mathrm{a}=\frac{2}{3}\ln \frac{2}{3}-\frac{1}{9}$

g(x) = ax

$\mathrm{g}\left(3\right)=3\left(\frac{2}{3}\ln \frac{2}{3}-\frac{1}{9}\right)$

= $\left(2\ln \frac{2}{3}-\frac{1}{3}\right)=\left(\ln \frac{4}{9}-\frac{1}{3}\right)=\ln \frac{4}{9}-\frac{1}{3}\mathrm{lne}=\ln \frac{4}{9}-{\mathrm{lne}}^{\frac{1}{3}}=\ln \frac{4}{9{\mathrm{e}}^{1/3}}={\log }_{\mathrm{e}}\frac{4}{9{\mathrm{e}}^{1/3}}$